模拟赛#1

Posted 2020-12-27 kaleidoscope233

虽然AK了但是手速不够快而且罚时爆炸QAQ....

A.Prime Ring Problem

找素数环,经典深搜问题,要注意找完排列再判断素数为超时,正确的做法是边搜边判断,注意输出格式==

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <queue>
 7 #include <map>
 8 #define ll long long
 9 #define out(a) printf("%d ",a)
10 #define writeln printf("
")
11 const int N=1e5+50;
12 using namespace std;
13 int n;
14 int a[N];
15 int tot=0,cnt=1;
16 bool flag,vis[N];
17 int read()
18 {
19     int s=0,t=1; char c;
20     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
21     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
22     return s*t;
23 }
24 ll readl()
25 {
26     int s=0,t=1; char c;
27     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
28     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
29     return s*t;
30 }
31 bool check(int x,int y)
32 {
33     int sum=x+y;
34     for (int i=2;i*i<=sum;i++)
35       if (sum%i==0) return false;
36     return true;
37 }
38 void dfs(int dep)
39 {
40     if (dep==n&&check(a[n],a[1])){
41         for (int i=1;i<=n;i++)
42           if (i!=n) out(a[i]);
43           else printf("%d",a[i]);
44         writeln;
45     }
46     for (int i=1;i<=n;i++){
47       if (!vis[i]&&check(a[cnt],i)){
48           a[++cnt]=i;
49           vis[i]=true;
50           dfs(dep+1);
51           cnt--;
52           vis[i]=false;
53       }
54     }
55 }
56 int main()
57 {
58     while (~scanf("%d",&n)){
59       printf("Case %d:
",++tot);
60       a[1]=1; vis[1]=true;
61       dfs(1); writeln;
62     }
63     return 0;
64 }

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B.Monkey and Banana

n个长方体,x,y,z可以作为长方体任意的长宽高,现在要你堆长方体,要求上面的长方体的长和宽严格小于下面的,求最大高度.

显然,长和宽是单调上升的,很容易想到排序后求LIS,由于求的是高度,可以把高度看作价值n2求一波LIS.

要注意的是长方体有6种放置方式,dp数组初值要设为第i个长方体高度(因为这个一直卡在这题...)

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <queue>
 7 #include <map>
 8 #define ll long long
 9 #define out(a) printf("%d ",a)
10 #define writeln printf("
")
11 const int N=1e5+50;
12 using namespace std;
13 int n,x,y,z,tot,cnt=0;
14 int ans;
15 int f[N];
16 struct node
17 {
18     int x,y,z,f;
19 }a[N];
20 int read()
21 {
22     int s=0,t=1; char c;
23     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
24     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
25     return s*t;
26 }
27 ll readl()
28 {
29     int s=0,t=1; char c;
30     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
31     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
32     return s*t;
33 }
34 bool cmp(node a,node b)
35 {
36     return a.x==b.x?a.y<b.y:a.x<b.x;
37 }
38 int main()
39 {
40     while (~scanf("%d",&n)!=0){
41       if (n==0) break;
42       ans=-233333; tot=0;
43       memset(f,0,sizeof(f));
44       memset(a,0,sizeof(a));
45       for (int i=1;i<=n;i++){
46         x=read(),y=read(),z=read();
47         a[++tot].x=x,a[tot].y=y,a[tot].z=a[tot].f=z;
48         a[++tot].x=x,a[tot].y=z,a[tot].z=a[tot].f=y;
49         a[++tot].x=y,a[tot].y=x,a[tot].z=a[tot].f=z;
50         a[++tot].x=y,a[tot].y=z,a[tot].z=a[tot].f=x;
51         a[++tot].x=z,a[tot].y=x,a[tot].z=a[tot].f=y;
52         a[++tot].x=z,a[tot].y=y,a[tot].z=a[tot].f=x;
53       }
54       sort(a+1,a+tot+1,cmp);
55       for (int i=2;i<=tot;i++){
56         for (int j=1;j<i;j++)
57           if (a[i].x>a[j].x&&a[i].y>a[j].y) a[i].f=max(a[i].f,a[j].f+a[i].z);
58         ans=max(ans,a[i].f);
59       }
60       printf("Case %d: maximum height = %d",++cnt,ans); writeln;
61     }
62     return 0;
63 }
64     

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C.最大报销额

题意就懒得说了,大概就是做一个带有小数的01背包。

把小数乘上100然后xjb写就可以了,要注意的是强制转换如:(double)x,当x为一个式子时最好加上括号(double)(x).

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cmath>
 6 #include <queue>
 7 #include <map>
 8 #define ll long long
 9 #define out(a) printf("%d ",a)
10 #define writeln printf("
")
11 const int N=3e6+50;
12 using namespace std;
13 double Q,suma,sumb,sumc,v;
14 int n,maxn,m,cnt=0;
15 int f[N],num[31];
16 bool flag;
17 char c;
18 int read()
19 {
20     int s=0,t=1; char c;
21     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
22     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
23     return s*t;
24 }
25 ll readl()
26 {
27     int s=0,t=1; char c;
28     while (c<‘0‘||c>‘9‘){if (c==‘-‘) t=-1; c=getchar();}
29     while (c>=‘0‘&&c<=‘9‘){s=s*10+c-‘0‘; c=getchar();}
30     return s*t;
31 }
32 bool check(int a,int b,int c)
33 {
34     int sum=a+b+c;
35     if (a<=600&&b<=600&&c<=600&&sum<=1000) return true;
36     return false;
37 }
38 int main()
39 {
40     while (~scanf("%lf%d",&Q,&n)){
41       if (n==0) break;
42       maxn=(int)(Q*100); cnt=0; 
43       memset(f,0,sizeof(f));
44       for (int i=1;i<=n;i++){
45         m=read(); flag=false;
46         suma=sumb=sumc=0;
47         for (int j=1;j<=m;j++){
48           scanf(" %c:%lf",&c,&v);
49           if (c==‘A‘) suma+=v;
50           else if (c==‘B‘) sumb+=v;
51           else if (c==‘C‘) sumc+=v;
52           else flag=true;
53          }
54           if (check(suma,sumb,sumc)&&!flag) num[++cnt]=(int)((suma+sumb+sumc)*100);
55       }
56           for (int i=1;i<=cnt;i++)
57             for (int j=maxn;j>=num[i];j--)
58               f[j]=max(f[j],f[j-num[i]]+num[i]);
59           printf("%.2lf",f[maxn]/100.0); writeln;
60       }
61     return 0;
62 }
63         

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